uva timecard

Test instructions: Morse code, enter a number of letters of the Morse number, a dictionary and several encodings. For each number, determine which word it might be,If more than one word matches exactly, output the first word and add a "! "; If you can't match exactly, add or remove as few characters as possible at the end of the code,Make it match a word and add "? ”。Analysis: The first time to do, a look at what ah, do not, now come back to see, found can be done, you can use the map of the STL

#include string>5#include 6#include 7#include 8 using namespacestd;9 Const intN =25000+5;Ten One stringS[n]; Avectorint>G[n]; -mapstring,int>M; - intDp[n]; the - stringChangestringSintPos//the delete operation and the change operation can be seen as the same - { - stringAns =""; + for(intI=0; I) - { + if(pos = = i) ans + ='*'; A ElseAns + =S[i]; at } - returnans; - } - - stringAddstringSintPOS) - { in stringAns =""; - for(intI=0; I) to { +

. Regardless of the variable ordering of the S-tree, the first character always describes the value of x1, the Seco nd character describes the value of x2, and so on. So, the line110Corresponds to the VVA ( x1 = 1, x2 = 1, x3 = 0).The input is terminated by a test case starting with n = 0. This test case is should not being processed.OutputFor each s-tree, output of the line 'S-tree #J:", whereJIs the number of the S-tree. Then print a line this contains the value of for each of the givenmVvas,

= Find (x), FY =Find (y); if(FX! = FY) Pre[fx] =fy; }BOOLisconnct ()//Determine whether the graph is connected, that is, all the points are in a set {intCNT =0; for(inti =1; I if((outdegree[i]! =0|| Indegree[i]! =0) Pre[i] = = i) cnt++; if(CNT = =1)return true; return false;}BOOLIseulur ()//whether there is Euler pathway {intCNT =0; intFlag =0; for(inti =1; I -; i++) if((outdegree[i]! =0|| Indegree[i]! =0) (indegree[i]! =Outdegree[i])) Judging the singularity, the method is n

#include #include#include#defineMAX 1010using namespacestd;DoublePL,PR;DoubleData[max];voidInit () {memset (data,0,sizeof(data));}DoubledpintN) { Doubletemp; intL,r; if(n==0) return 0; if(data[n]>0) returnData[n]; for(intI=1; i) {L=i-1; R=n-i; Temp= DP (l) +DP (R) + (DP (l) *PL+DP (R) *pr+1)/(1-pl-PR); if(i==1) Data[n]=temp; Else if(Temp Data[n]) Data[n]=temp; } returndata[n];}intMain () {intN; while(SCANF ("%D%LF%LF", AMP;N,AMP;PL,AMP;PR)!=eof n!=0) {Init (); printf ("%.2

"Topic link": Click here~~"The main idea": Euler function: The number of the number of N-"Thinking": Bare Euler function, pay attention to the case of N==1, n==1 case, should output 0,POJ still judge 1 can also, but old Ojuva must be 0 before, pay attention to.Code:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. POJ 2407 Relatives UVA 10299 relatives (Euler function)

D[u][0][1]=sum{d[v][0]}D[u][1][1]=d[u][1][0]=sum{min (d[v][0],d[v][1])}D[u][0][0]=sum{d[v][0]}-d[min][0]+d[min][1] (D[min][1]=min (d[v][1]))It's been a long time. See LRJ is the first DFS to find a good path, and then the reverse direction along the path recursion. O (n).Inefficient lately!Who laughs last who laughs best!1#include 2#include 3#include 4#include 5 #definePB Push_back6 using namespacestd;7 Const intinf=1e4;8 Const intmaxn=10010;9 intN;Tenvectorint>VT[MAXN]; One intd[maxn][2][2]; A

The main idea: a ring can open and close. Now there are N (1Title Analysis: The set of rings to be opened with a binary number, a total of 2^n cases, enumeration of each case. When the ring is opened, the ring is isolated, the next is to judge the rest of the ring is connected with a few rings, if some ring is still connected with more than two rings, the scheme is not feasible, it is impossible to form a chain, and then determine whether the remaining ring is connected to a circle, if there is,

Test instructions: There is a bee, every year the drone produces a females and drone, and then die, females produce a drone, and then die, there is not a dead females, ask the number of drones and the total number of bees after n yearsIdea: recursion + test InstructionsThe number of drones this year is m[x], last year was m[x-1],The number of females this year is f[x], last year for F[x-1]According to test instructions, you can get:M[X]=F[X-1]+M[X-1];F[x]=m[x-1]+1PS: According to recursion can a

Title: To an nTopic Analysis: First of all to determine whether from 1 to N, and then backtracking.Note: This problem has a pit, according to the sample output will be PE.The code is as follows:# include　　UVA-208 Firetruck (backtracking)

the same).InputThe input consists of T test Cases. The number of test cases T is given on the first line of the input file. Each test case takes one line containing a circular sequence so is written as an arbitrary linear sequence. Since the circular sequences is DNA sequences, only four symbols, A, C, G and T, is Allowed. Each sequence have length at least 2 and at the most.OutputPrint exactly one line for each test case. The line was to contain the lexicographically smallest sequence for t

Title: Give a string of parentheses to see if it matches.Topic Analysis: A beginning with the interval DP write, timed out ...Note: The empty string is valid.The code is as follows:# include　　UVA-673 parentheses Balance (stack)

; WEI[B]|=Wei[a]; }}voidInituf () { for(inti =0; I ) {Pa[i]=i; Wei[i]=1i; }}mapint,Double>Meo;mapint,Double>:: iterator it;#defineMP Make_pair#defineFi first#defineSe SecondDoubletr;Doubledpints) { if((it = Meo.find (s))! = Meo.end ())returnIt->second; DoubleAns =0; intLKS =0; for(inti = n; i--;){ if(s>>i1) {lks++;Continue; } Ans+ = DP (s|Wei[i]); } ans= (n1.) /(n-lks) * (Tr*ans +1); Meo.insert (MP (S,ans)); returnans;}//#define LOCALintMain () {#ifdef LOCAL freopen ("In.txt","R",

"Source" https://uva.onlinejudge.org/index.php?option=com_onlinejudgeitemid=8page=show_problem problem=2266This problem is relatively simple, the main work is the segmentation of the string, how to write a beautiful code point is the focus, readability is very important. The use of function atof (stdlib.h) is more convenient than a bit to spell numbers.#include #include#include#include#defineMaxLen 100005using namespacestd;intSubexplen (Char* S,intstart) { intlen=0; if(s[start]!='(') {

Idea: Simple EnumerationAC Code:#include Uva-10167-birthday Cake (Simple enumeration)

Determine if there is a Euler loop as long as two conditionsDiagram is connected, no singularity points existPay attention to the case where the margin is 0. In addition this problem data pits.#include #include #include using namespace Std;const int maxn=208;struct fuck{int u,v,next;}EDGE[MAXN*MAXN];int HEAD[MAXN];int tol;void Init (){tol=0;memset (head,-1,sizeof (head));}void Addedge (int u,int v){Edge[tol].u=u;Edge[tol].v=v;Edge[tol].next=head[u];head[u]=tol++;}BOOL VIS[MAXN];int DU[MAXN];void

#include #defineREP (I,A,B) for (int i=a;i#defineMS0 (a) memset (A,0,sizeof (a))using namespaceStd;typedefLong Longll;Const intmaxn=2000100;Const intinf=1 in;intS,n;intX[MAXN];intCNT[MAXN],M[MAXN];intMain () {Freopen ("In.txt","R", stdin); Freopen ("OUT.txt","W", stdout); intT;cin>>u; while(t--) {scanf ("%d%d",s,N); REP (i,1, N) scanf ("%d",X[i]); MS0 (CNT); MS0 (m); intk=1; intans=0, tag=1; REP (i,1, N) {Cnt[x[i]+k*s]++; if(cnt[x[i]+k*s]==2) m[k]++,tag=0; if(i%s==0) k++; } if(tag

Test instructionsGiven the set of 4 N (1 Analysis:Obviously the four-heavy cycle is not enough, I first thought is to use a map to save A+b,c+d, and then in the search statistics. Timeout....And then the book said with a hash table to achieve, see some of the hash of the puzzle is too ingenious, learn a bit.There is the problem can be solved with two points, first calculate the A+b, and then enumerate C+d, and then two points to find the scope.Hash 630ms:#include #include Constintn=4005;inta[4][

https://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8page=show_problemproblem= 565A record path can use a two-dimensional array to record the amount of change. Then you can push all the values forward from the back.#include #includestring>#include#include#include#include#include#includeusing namespacestd;#defineMEM (A, B) memset (A,b,sizeof (a))#definePF printf#defineSF scanf#defineDebug printf ("!\n")#defineINF 8000#defineMAX (A, b) a>b?a:b#defineBlank pf ("\ n")#defineLL Long Lo

Topic PortalTest Instructions: Training guide P189Analysis: The idea of a complete reference book, K^k table into an orderly table:Table 1:a1 + B1 Table 2:a2 + B1 .......Table K:ak + B1 can maintain a K-length array representing the current top-K small number and, when the array of line I is read, First push first, that is the smallest, then can be updated to the second, that is-b[i] + b[i+1]. The logn of the priority queue can be used to get the minimum value of each moment, and the total co